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Let $g:\mathbb{R}^n\times \mathbb{R}^{n'}\times [0,1]\rightarrow \mathbb{R}^n$ a continuously differentiable function. Can we deduce that $g$ is locally Lipschitz ?

which is equivalent that for all nonempty compact sets $K\subset\mathbb{R}^n\times \mathbb{R}^{n'}\times [0,1] $, there exosts $M>0$ such that $$\| g(x_1,y_1,t)-g(x_2,y_2,t) \| \leq M(\|x_1-x_2 \| + \|y_1-y_2 \| ),$$ for all $(x_1,y_1,t),(x_2,y_2,t) \in K$ ?

Or should we assume some additional convexity hypothesis ?

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  • $\begingroup$ No, you don't need an additional convexity hypothesis. You can always enlarge $K$ to a convex compact set. $\endgroup$
    – MaoWao
    Jun 22, 2021 at 8:58

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Yes, we can already deduce local Lipschitzianity. Just note that $$ \sup_{(x, y, t) \in \overline{B} \times J} \lVert Dg(x, y, t) \rVert < \infty $$ for some ball $B \subseteq \mathbb{R}^n \times \mathbb{R}^{n'}$ and interval $J$ such that $B \times J \supseteq K$ because of domain compactness and continuity of $Dg$. We can choose such ball and interval because $K$ is bounded.

Then use the Mean Value Theorem to follow Lipschitz-continuity on $\overline{B} \times J$ which implies Lipschitz-continuity on $K$.

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  • $\begingroup$ just to clarify, the (Cauchy) mean-value theorem you link to is the one for single-variable calculus, and that is not applicable here (the standard single-variable MVT fails badly in higher dimensions). What we do have is the mean-value inequality (same page, just a little further down... though that still has the strong, and unnecessary hypothesis of continuous differentiability) which is valid for any differentiable map between Banach spaces. $\endgroup$
    – peek-a-boo
    Jun 22, 2021 at 9:44
  • $\begingroup$ Oh sorry, wrong link $\endgroup$
    – Meowdog
    Jun 22, 2021 at 9:49

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